The time taken by a block of wood (initially | vedclass

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(B) Given: Distance $S = 9.8 \ m$,initial velocity $u = 0$,angle of inclination $	heta = 30^o$,acceleration due to gravity $g = 9.8 \ m/s^2$.For a bl

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$2$

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(B) Given: Distance $S = 9.8 \ m$,initial velocity $u = 0$,angle of inclination $	heta = 30^o$,acceleration due to gravity $g = 9.8 \ m/s^2$.For a block sliding down a smooth inclined plane,the acceleration $a = g \sin 	heta$.Using the equation of motion $S = ut + \frac{1}{2}at^2$:$9.8 = 0 \cdot t + \frac{1}{2} (g \sin 30^o) t^2$$9.8 = \frac{1}{2} 	imes 9.8 	imes \sin 30^o 	imes t^2$$9.8 = \frac{1}{2} 	imes 9.8 	imes 0.5 	imes t^2$$1 = 0.25 	imes t^2$$t^2 = \frac{1}{0.25} = 4$$t = 2 \ sec$.

The time taken by a block of wood (initially at rest) to slide down a smooth inclined plane $9.8 \ m$ long (angle of inclination is $30^o$) is ......... $sec$.

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